Ganita Prakash · Grade 6 · Maths

Chapter 2 — Lines and Angles

Complete, step-by-step solutions to every question with diagrams & explanations.

Points · Line Segments · Lines · Rays · Angles · Measuring & Drawing Angles

💡In-text QuestionsExplore · Think! · Let's Explore 📝Exercise QuestionsAll "Figure it Out" solutions

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Section 2.4 Section 2.5 Section 2.6 Section 2.8 Section 2.9 Where are the Angles? (page 45)Section 2.10 Section 2.11

💡 In-text Questions

Answers to the exploratory prompts, "Think!" boxes and "Let's Explore" questions inside the chapter.

(Section 2.6, Comparing Angles) Is it always easy to compare two angles?

Solution

No — it is not always easy to compare two angles just by looking. When two angles are very close in size, the eye cannot judge which is bigger.

For example, a \(89^\circ\) angle and a \(91^\circ\) angle look almost identical, so they cannot be compared reliably without measuring or overlapping (superimposition). For angles that differ a lot, a direct look is enough.

Answer

It is not always easy. Angles like \(89^\circ\) and \(91^\circ\) need measurement or superimposition to compare; clearly different angles can be compared by sight.

(Section 2.6) Where else do we use superimposition to compare?

Solution

Superimposition means placing one figure exactly over another to see which is larger. We use the same idea to compare the sizes of many things, not just angles.

Answer

A few examples are line segments (place one over the other to see which is longer), squares and circles. Think of more such shapes!

(Section 2.8, Special Types of Angles) Is it possible to draw ray \(\overrightarrow{OC}\) such that the two angles are equal to each other in size?

Solution

Take the straight angle \(\angle AOB\) marked on the edge of a rectangular paper and fold it so that ray \(OB\) falls exactly on ray \(OA\).

Answer

Yes. When \(OA\) and \(OB\) overlap on folding, the crease \(OC\) divides \(\angle AOB\) into two equal angles \(\angle AOC\) and \(\angle COB\).

(Section 2.8) If a straight angle is formed by half of a full turn, how much of a full turn will form a right angle?

Solution

A straight angle \(=\tfrac12\) of a full turn. A right angle is exactly half of a straight angle.

\[\text{Right angle}=\frac12\times\left(\frac12 \text{ turn}\right)=\frac14\ \text{turn}\]

Answer

A right angle is \(\dfrac{1}{4}\) of a full turn.

(Section 2.9, Make your own protractor) A circle (full turn = \(360^\circ\)) is divided into \(1,2,3,4,5,6,8,9,10\) and \(12\) equal parts. What are the degree measures of the resulting angles?

Solution

Each equal part \(=\dfrac{360^\circ}{n}\), where \(n\) is the number of parts.

Parts (n)	1	2	3	4	5	6	8	9	10	12
Each angle	360°	180°	120°	90°	72°	60°	45°	40°	36°	30°

For example \(\dfrac{360^\circ}{5}=72^\circ\) and \(\dfrac{360^\circ}{8}=45^\circ\).

(Section 2.9, labelled protractor — page 36) Name the different angles in the figure and write their measures.

Rays P, Q, R, S, T, U on the protractor (centre O)

Solution

Reading each ray from the protractor and finding the difference of the two readings gives each angle:

Angle	Measure	Angle	Measure
∠POQ	35°	∠QOT	125°
∠POR	95°	∠QOU	145°
∠POS	125°	∠ROS	30°
∠POT	160°	∠ROT	65°
∠QOR	60°	∠ROU	85°
∠QOS	90°	∠SOT	35°
∠SOU	55°	∠TOU	20°

(Section 2.9, Think! — page 40) In Fig. 2.19, \(\angle AOB=\angle BOC=\angle COD=\angle DOE=\angle EOF=\angle FOG=\angle GOH=\angle HOI = \underline{\ \ ?\ \ }\). Why?

Solution

The straight angle of \(180^\circ\) was folded into halves again and again until it was split into 8 equal parts. So each small angle is

\[\frac{180^\circ}{8}=22.5^\circ\]

Answer

Each angle \(=22.5^\circ\), because the straight angle \(180^\circ\) is divided into 8 equal parts.

(Section 2.11, Let's Explore — page 53) In the figure, \(\angle TER = 80^\circ\). What is the measure of \(\angle BET\)? What is the measure of \(\angle SET\)?

∠TER = 80°, ∠SER = 90°, with B–E–R a straight line

Solution

Finding \(\angle BET\): Points \(B, E, R\) lie on a straight line, so \(\angle BER\) is a straight angle \(=180^\circ\). This is split into \(\angle BET\) and \(\angle TER\):

\[\angle BET = 180^\circ - \angle TER = 180^\circ - 80^\circ = 100^\circ\]

Finding \(\angle SET\): \(ES\) is perpendicular to \(ER\), so \(\angle SER = 90^\circ\). Now \(\angle SER\) is made of \(\angle SET\) and \(\angle TER\):

\[\angle SET = \angle SER - \angle TER = 90^\circ - 80^\circ = 10^\circ\]

Answer

\(\angle BET = 100^\circ\) and \(\angle SET = 10^\circ\).

📝 Exercise Questions

Detailed solutions to all the "Figure it Out" exercises, section by section.

Section 2.4 — Ray

Can you help Rihan and Sheetal find their answers?

Rihan marked a point on a piece of paper. How many lines can he draw that pass through the point?

Sheetal marked two points on a piece of paper. How many different lines can she draw that pass through both points?

Solution

Rihan (one point): Through a single point we can draw a line in any direction — turning it a little always gives a new line. So there is no limit.

Sheetal (two points): Two points fix a straight path completely, so only one straight line can pass through both.

Answer

Rihan can draw infinitely many (uncountable) lines through the point. Sheetal can draw only one line through the two points.

Name the line segments in Fig. 2.4. Which of the five marked points are on exactly one of the line segments? Which are on two of the line segments?

Solution

Reading the joined points along the path L–M–P–Q–R gives four line segments. An end point of the whole path lies on only one segment; a turning point in the middle lies on two segments.

Answer

Line segments: LM, MP, PQ, QR.
Points on exactly one line segment: L and R (the ends).
Points on two line segments: M, P and Q.

Name the rays shown in Fig. 2.5. Is T the starting point of each of these rays?

Solution

A ray is named by its starting point first, then any other point it passes through.

Answer

Rays: \(\overrightarrow{TA},\ \overrightarrow{TB},\ \overrightarrow{TN}\) and \(\overrightarrow{NB}\).
No. T is the starting point of \(\overrightarrow{TA},\overrightarrow{TB},\overrightarrow{TN}\), but not of \(\overrightarrow{NB}\) (that ray starts at N).

Draw a rough figure and write labels appropriately to illustrate each of the following:

a. \(\overleftrightarrow{OP}\) and \(\overleftrightarrow{OQ}\) meet at O. b. \(\overrightarrow{XY}\) and \(\overleftrightarrow{PQ}\) intersect at point M.
c. Line \(l\) contains points E and F but not point D. d. Point P lies on \(\overline{AB}\).

Solution

Each rough sketch simply shows the relationship described:

(a) lines meeting at O (b) intersecting at M (c) line l through E, F (not D) (d) P on AB

In Fig. 2.6, name: a. Five points b. A line c. Four rays d. Five line segments

Solution

All of D, E, O, B, C lie along one straight line (with O also the start of two more rays).

Answer

a. Five points: D, E, O, B, C
b. A line: \(\overleftrightarrow{DE}\) (also \(\overleftrightarrow{DO},\overleftrightarrow{DB},\overleftrightarrow{EO},\overleftrightarrow{EB},\overleftrightarrow{OB}\))
c. Four rays: \(\overrightarrow{OC},\overrightarrow{OB},\overrightarrow{OE},\overrightarrow{OD}\) (others are possible)
d. Five line segments: \(\overline{DE},\overline{DO},\overline{DB},\overline{EO},\overline{EB}\) (\(\overline{OB},\overline{OC}\) also possible)

Here is a ray \(\overrightarrow{OA}\) (Fig. 2.7). It starts at O and passes through the point A. It also passes through the point B.

a. Can you also name it as \(\overrightarrow{OB}\)? Why? b. Can we write \(\overrightarrow{OA}\) as \(\overrightarrow{AO}\)? Why or why not?

Solution

A ray is fixed by its starting point and the direction it travels.

Answer

a. Yes. O is the starting point and B lies on the same ray going in the direction of A. So \(\overrightarrow{OA}\) and \(\overrightarrow{OB}\) are the same ray (\(\overrightarrow{OA}\) is just the extension of \(\overrightarrow{OB}\)).
b. No. \(\overrightarrow{OA}\) starts at O, while \(\overrightarrow{AO}\) starts at A. Different starting points mean different rays.

Section 2.5 — Angle

Can you find the angles in the given pictures? Draw the rays forming any one of the angles and name the vertex of the angle.

Frame of the bicycle (points A, B, C, D)

Solution

Wherever two rods (rays) meet, an angle is formed. In the bicycle frame, two rods meet at D.

Answer

Yes. One of the angles is \(\angle BDC\). Its vertex is D; one ray is \(\overrightarrow{DC}\) and the other ray is \(\overrightarrow{DB}\). (Try to find angles in the other pictures too.)

Draw and label an angle with arms \(ST\) and \(SR\).

Solution

The two arms both start from the common point S, which becomes the vertex. One arm goes to T and the other to R.

∠TSR (or ∠RST) with vertex S and arms ST, SR

Explain why \(\angle APB\) cannot be labelled as \(\angle P\).

Solution

At the point P, three different rays (\(\overrightarrow{PA}, \overrightarrow{PB}, \overrightarrow{PC}\)) start out. So several angles share the same vertex P — namely \(\angle APB\), \(\angle BPC\) and \(\angle APC\).

Answer

If we wrote just \(\angle P\), we would not know which pair of arms is meant, because more than one angle has vertex P. Naming it with three letters, \(\angle APB\) (vertex in the middle), removes this confusion. So it cannot simply be called \(\angle P\).

Name the angles marked in the given figure.

Rays TP, TQ, TR from vertex T (two angles marked)

Solution

Both marked angles have vertex T, measured from arm \(\overrightarrow{TR}\).

Answer

The marked angles are \(\angle RTQ\) and \(\angle RTP\).

Mark any three points (not on one line) as A, B, C. Draw all possible lines through pairs. How many lines do you get? How many angles can you name using A, B, C?

Solution

With 3 non-collinear points, each pair gives one line: (A,B), (B,C), (C,A).

Three lines AB, BC, CA and the angles at A, B, C

Answer

We get 3 lines: \(\overleftrightarrow{AB},\ \overleftrightarrow{BC},\ \overleftrightarrow{CA}\).
We can name 3 angles: \(\angle ABC\ (=\angle CBA)\), \(\angle BCA\ (=\angle ACB)\) and \(\angle CAB\ (=\angle BAC)\).

Mark any four points so that no three are on one line, as A, B, C, D. Draw all possible lines through pairs. How many lines? How many angles can you name using A, B, C, D?

Solution

With 4 points, the number of lines is the number of pairs \(=\dbinom{4}{2}=6\).

Six lines and the angles formed at A, B, C, D

Answer

We get 6 lines: \(\overleftrightarrow{AB},\overleftrightarrow{BC},\overleftrightarrow{CD},\overleftrightarrow{DA},\overleftrightarrow{AC},\overleftrightarrow{BD}\).
We can name 12 angles: \(\angle BAC,\angle CAD,\angle BAD,\angle ADB,\angle BDC,\angle ADC,\angle DCA,\angle ACB,\angle DCB,\angle CBD,\angle DBA,\angle CBA\).

Section 2.6 — Comparing Angles

Fold a rectangular sheet of paper, then draw a line along the fold. Name and compare the angles formed between the fold and the sides of the paper. Which is the largest and smallest angle you made?

Solution

The slanting fold line meets the top side at F and the bottom side at E, making four angles.

Fold line EF making ∠AEF, ∠BEF, ∠CFE, ∠DFE

Answer

Angles formed: \(\angle AEF,\ \angle BEF,\ \angle DFE,\ \angle CFE\).
Here \(\angle AEF\) and \(\angle CFE\) are the larger angles, while \(\angle BEF\) and \(\angle DFE\) are the smaller ones. (Fold in different ways to make different largest/smallest angles.)

In each case, determine which angle is greater and why: a. \(\angle AOB\) or \(\angle XOY\) b. \(\angle AOB\) or \(\angle XOB\) c. \(\angle XOB\) or \(\angle XOC\).

Solution

Compare by seeing whether one angle contains the other (a whole is bigger than its part).

Answer

a. \(\angle AOB\) is greater — \(\angle XOY\) is only a part of it, since \(\angle AOB=\angle AOX+\angle XOY+\angle YOB\).
b. \(\angle AOB\) is greater.
c. Neither — \(\angle XOB=\angle XOC\) (they are equal).

Which angle is greater: \(\angle XOY\) or \(\angle AOB\)? Give reasons.

Solution

Here neither angle is a part of the other and they look close in size.

Answer

By only looking at the figure, we cannot tell which is greater. We must use superimposition (trace one and place it over the other) or measure them with a protractor.

Section 2.8 — Special Types & Classifying Angles

How many right angles do the windows of your classroom contain? Do you see other right angles in your classroom?

Solution

Each rectangular window pane has square corners.

Answer

A rectangular window has 4 right angles (one at each corner). Other right angles in the classroom: corners of the door, blackboard, tables, tiles, books, etc.

Join A to other grid points by a straight line to get a straight angle. What are all the different ways of doing it?

Solution

A straight angle at A needs the two chosen points to be on the same straight line as A, one on each side (opposite directions), so the two rays open out to \(180^\circ\).

Answer

Draw a straight line right through A with grid points on both sides. This is possible along the horizontal, the vertical, and the two diagonal directions — and along any other slope that has grid points on both sides of A. Every such line gives a straight angle at A.

Now join A to other grid points to get a right angle. What are all the different ways of doing it? (Hint: extend the line so that a line through A divides the straight angle CAB into two equal parts.)

s28_q3_grid — Making a right angle at A (and the hint of bisecting a straight angle)

s28_q3_hint — Making a right angle at A (and the hint of bisecting a straight angle)

Solution

A right angle is exactly half of a straight angle. So at A we pick two directions that are perpendicular to each other.

Answer

Pick two rays through A at right angles, for example:
• horizontal and vertical,
• the two diagonals (they are perpendicular to each other),
• or a slope like “1 up, 2 right” together with its perpendicular “2 up, 1 left”.
Each perpendicular pair gives a right angle at A.

Get a slanting crease on the paper. Now try to get another crease perpendicular to the slanting crease.
a. How many right angles do you have now? Justify why they are exact right angles.
b. Describe how you folded the paper so that anyone can follow.

Solution

Fold once to make the slanting crease. Fold again so the crease falls exactly on itself — the new crease is perpendicular to it.

Answer

a. You get 4 right angles. Each is exactly \(\tfrac14\) of the full turn around the crossing point. Because the fold makes the two halves coincide (superimpose), the four angles are proved equal, so each \(=\dfrac{360^\circ}{4}=90^\circ\).
b. How to fold: make a straight crease; then fold the paper so that this crease lies exactly on top of itself. Press and unfold — the second crease crosses the first at right angles.

Classify 2

Make a few acute angles and a few obtuse angles. Draw them in different orientations.

Solution

Any angle smaller than a right angle is acute; any angle bigger than a right angle but smaller than a straight angle is obtuse. Draw several pointing in different directions (up, down, sideways, slanting).

Answer

Sample acute angles (small opening) and obtuse angles (wide opening) can be drawn in many orientations — the orientation does not change the type; only the amount of opening does.

Classify 3

Acute means sharp and obtuse means blunt. Why do you think these words have been chosen?

Solution

Look at the opening between the two arms.

Answer

In an acute angle the arms open only a little, making a narrow, sharp point — like a sharp tip. In an obtuse angle the arms open wide, giving a blunt (broad) corner. That is why they are named ‘sharp’ and ‘blunt’.

Classify 4

Find out the number of acute angles in each of the figures below. What will be the next figure and how many acute angles will it have? Do you notice any pattern?

Solution

Count the acute angles inside each figure:

Figure (i): \(3\) Figure (ii): \(12\) Figure (iii): \(21\)

The counts increase by \(9\) each time: \(3,\ 12,\ 21,\ \ldots\) So the next figure has \(21+9=30\).

A neat formula for the \(n\)-th figure is \(9n-6\): check \(9(1)-6=3,\ 9(2)-6=12,\ 9(3)-6=21,\ 9(4)-6=30\). (The extra triangles keep adding groups of angles as more inner triangles appear.)

Answer

(i) 3 (ii) 12 (iii) 21. The next figure will have 30 acute angles. Pattern: each figure has 9 more acute angles than the previous one (formula \(9n-6\)).

Section 2.9 — Measuring Angles

1 (a-c)

Write the measures of the following angles from the protractor: a. \(\angle KAL\) b. \(\angle WAL\) c. \(\angle TAK\).

Unlabelled protractor with rays K, L, W, T (vertex A)

Solution

The vertex A is at the centre of the protractor, so we count the number of \(1^\circ\) units between the two arms (counting in 5s and 10s using the medium and long marks).

Answer

a. \(\angle KAL = 30^\circ\) b. \(\angle WAL = 50^\circ\) c. \(\angle TAK = 120^\circ\).
Yes — the long marks (every \(10^\circ\)) and medium marks (every \(5^\circ\)) let us count quickly in 5s or 10s.

1 (p40)

Find the degree measures of the following angles using your protractor.

Solution

Place the centre of the protractor on the vertex H and read each angle.

Answer

First figure: \(\angle IHJ = 47^\circ\) Middle figure: \(23^\circ\) Right figure: \(108^\circ\).

Find the degree measures for the angles given below. Check if your paper protractor can be used here!

Solution

Measure with the standard protractor.

Answer

\(\angle IHJ = 42^\circ\) (left) and \(\angle IHJ = 116^\circ\) (right).
No, the handmade paper protractor cannot be used here — its creases only fall at multiples of \(22.5^\circ\) (\(0,22.5,45,67.5,90,\ldots\)), so it cannot measure \(42^\circ\) or \(116^\circ\) exactly.

How can you find the degree measure of the angle given below using a protractor?

Solution

A protractor only measures up to \(180^\circ\). For a reflex angle, first measure the smaller (unmarked) angle, then subtract from a full turn:

\[\text{Reflex angle}=360^\circ-\text{(smaller angle)}=360^\circ-100^\circ=260^\circ\]

Answer

Measure the smaller angle (\(100^\circ\)) and subtract from \(360^\circ\): the marked reflex angle \(=260^\circ\).

Measure and write the degree measures for each of the following angles (a–f).

Solution

Placing the protractor centre on each vertex and reading the scale gives:

Answer

a. \(80^\circ\) b. \(120^\circ\) c. \(60^\circ\) d. \(130^\circ\) e. \(130^\circ\) f. \(60^\circ\).

Find the degree measures of \(\angle BXE, \angle CXE, \angle AXB\) and \(\angle BXC\).

Rays A, B, C, E from vertex X on the protractor

Solution

Read the ray positions from the protractor. Note \(A\)–\(X\)–\(E\) is a straight line (\(180^\circ\)).

\(\angle BXE = 115^\circ\); \(\angle CXE = 85^\circ\).
Then \(\angle AXB = 180^\circ - 115^\circ = 65^\circ\) and \(\angle BXC = \angle BXE - \angle CXE = 115^\circ - 85^\circ = 30^\circ\).

Answer

\(\angle BXE = 115^\circ,\ \angle CXE = 85^\circ,\ \angle AXB = 65^\circ,\ \angle BXC = 30^\circ\).

Find the degree measures of \(\angle PQR, \angle PQS\) and \(\angle PQT\).

Solution

Measure each angle from arm \(\overrightarrow{QP}\).

Answer

\(\angle PQR = 45^\circ,\ \angle PQS = 100^\circ,\ \angle PQT = 150^\circ\).

Measure all three angles of each triangle in Fig. 2.21, and add them up. What do you get? Make a conjecture.

Solution

On measuring and adding the three angles of each triangle:

\[\angle A + \angle B + \angle C = 180^\circ\]

The same total \(180^\circ\) comes up for every triangle you try.

Answer

The three angles of a triangle always add up to \(180^\circ\). Conjecture: the sum of the angles of any triangle is a straight angle (\(180^\circ\)).

Where are the Angles? (page 45)

Angles in a clock: a. At 1 o'clock the angle between the hands is \(30^\circ\). Why? b. What will be the angle at 2 o'clock? 4 o'clock? 6 o'clock? c. Explore other angles.

Solution

The full turn at the centre (\(360^\circ\)) is split by the 12 hour-marks into 12 equal parts:

\[\text{Angle between two numbers}=\frac{360^\circ}{12}=30^\circ\]

a. At 1 o'clock the hands are 1 part apart \(=30^\circ\).
b. \(2\ \text{o'clock}=2\times30^\circ=60^\circ\); \(4\ \text{o'clock}=4\times30^\circ=120^\circ\); \(6\ \text{o'clock}=6\times30^\circ=180^\circ\).
c. e.g. \(3\ \text{o'clock}=90^\circ\), \(9\ \text{o'clock}=270^\circ\) (or \(90^\circ\) the other way).

Answer

Because \(360^\circ\div12=30^\circ\). So \(2\to60^\circ,\ 4\to120^\circ,\ 6\to180^\circ\), and (for example) \(3\to90^\circ,\ 9\to270^\circ\).

The angle of a door: Is it possible to express the amount by which a door is opened using an angle? What will be the vertex and the arms?

Solution

Yes. As a door swings open, it turns about its hinge.

Answer

Yes. The vertex is the hinge line (where the door meets the wall/frame). The arms are the edge of the wall and the edge of the door. The angle between them shows how much the door is opened.

Vidya on a swing: the greater the starting angle, the greater the speed. But where is the angle?

Solution

Think of the rope's rest position and its highest position.

Answer

The angle is between the rope's rest (starting) position and the rope's position at the highest point of the swing. Vertex = the point where the rope is tied to the branch; arms = the rope at rest and the rope at its highest point.

A toy with slanting slabs: the greater the slope, the faster the balls roll. Can angles describe the slopes? What are the arms? Which arm is visible and which is not?

Solution

A steeper slab makes a bigger angle with the horizontal, so the ball rolls faster.

Answer

Yes — a larger angle means a steeper slope. For each slab, one arm is the horizontal line (base) and the other arm is the slanting edge of the slab. Usually the horizontal arm is not drawn (invisible), while the slanting edge is visible.

An insect and its rotated version: Can angles describe the amount of rotation? What are the arms and the vertex? (Hint: observe the horizontal line touching the insects.)

Solution

The horizontal line touching both insects acts as the fixed starting arm.

Answer

Yes. One arm is the horizontal line through the insect (its starting position); the other arm is the line along the rotated insect. The vertex is the point about which it turns, and the angle between the arms equals the amount of rotation.

Section 2.10 — Drawing Angles

In Fig. 2.23, list all the angles possible. Guess their measures, then measure with a protractor and compare.

Solution

Wherever two lines cross or meet, angles are formed. List every such angle, guess it, then check with a protractor.

Answer

Some of the angles are: \(\angle CAP,\ \angle ACD,\ \angle APL,\ \angle DLP,\ \angle RPL,\ \angle SLP,\ \angle PRS,\ \angle LSR,\ \angle BRS,\ \angle CLP\) (and more). Record your guesses and measured values in a table and see how close they are.

Use a protractor to draw angles having the following degree measures: a. \(110^\circ\) b. \(40^\circ\) c. \(75^\circ\) d. \(112^\circ\) e. \(134^\circ\).

Solution

Steps for each angle: (1) draw a base ray; (2) place the protractor centre on its starting point and align the base with the \(0^\circ\) line; (3) mark the given degree; (4) join to form the angle.

Answer

Draw each angle by counting up to the required degree on the protractor and joining the mark to the vertex — giving \(110^\circ, 40^\circ, 75^\circ, 112^\circ\) and \(134^\circ\) angles.

Section 2.11 — Types of Angles & their Measures

In each grid, join A to another grid point by a straight line to get: a. an acute angle b. an obtuse angle c. a reflex angle. Mark the intended angles with curves.

Solution

Keep one ray fixed (say horizontal) and pick the second ray so that the opening is:

Answer

• Acute — opening less than a right angle (\(<90^\circ\)).
• Obtuse — opening between a right angle and a straight angle (\(90^\circ\!-\!180^\circ\)).
• Reflex — opening measured the long way round, more than a straight angle (\(>180^\circ\)).
Mark each intended angle with a small curve so it is clear which angle you mean.

Use a protractor to find the measure of each angle. Then classify each as acute, obtuse, right or reflex: a. \(\angle PTR\) b. \(\angle PTQ\) c. \(\angle PTW\) d. \(\angle WTP\).

Solution

Measure each from the correct arm. Note that \(\angle PTW\) and \(\angle WTP\) are the two ways round the same pair of arms, so they add to \(360^\circ\): \(102^\circ+258^\circ=360^\circ\).

Answer

a. \(\angle PTR = 30^\circ\) — acute
b. \(\angle PTQ = 60^\circ\) — acute
c. \(\angle PTW = 102^\circ\) — obtuse
d. \(\angle WTP = 258^\circ\) — reflex

Draw

Draw angles with the following degree measures: a. \(140^\circ\) b. \(82^\circ\) c. \(195^\circ\) d. \(70^\circ\) e. \(35^\circ\). Then estimate & classify each.

Solution

Use the protractor to draw each angle from a base ray. For the reflex angle \(195^\circ\), draw the smaller \(360^\circ-195^\circ=165^\circ\) part first and mark the reflex side.

Answer

\(140^\circ\) & \(82^\circ\) are obtuse; \(70^\circ\) & \(35^\circ\) are acute; \(195^\circ\) is a reflex angle.

Estimate

Estimate the size of each angle below, then measure it with a protractor and classify it as acute, right, obtuse or reflex.

Solution

How to classify at a glance: compare each opening with a right angle (\(90^\circ\)) and a straight angle (\(180^\circ\)):

Answer

• Less than \(90^\circ\) → acute • exactly \(90^\circ\) → right
• between \(90^\circ\) and \(180^\circ\) → obtuse • more than \(180^\circ\) → reflex
Angle (f) is drawn with a full round curve, so it is a reflex angle. First guess each value, then check with your protractor and see how close your guess was.

Make any figure with three acute angles, one right angle and two obtuse angles.

Solution

A ‘crown’ shape works nicely:

A crown: acute at A, B, C; right angle at E; obtuse at D, F

Answer

In this crown, \(\angle A,\ \angle B,\ \angle C\) are the three acute angles, \(\angle E\) is the right angle, and \(\angle D,\ \angle F\) are the two obtuse angles.

Draw the letter ‘M’ such that the angles on the sides are \(40^\circ\) each and the angle in the middle is \(60^\circ\).

Solution

Two slanting strokes make the sides (\(40^\circ\) each) and the middle V gives \(60^\circ\).

Draw the letter ‘Y’ such that the three angles formed are \(150^\circ, 60^\circ\) and \(150^\circ\).

Solution

The two upper arms make \(60^\circ\) between them, and each arm makes \(150^\circ\) with the stem. Check: \(150^\circ+60^\circ+150^\circ=360^\circ\) (a full turn about the joining point).

Letter Y with 60° at top and 150° on each side

The Ashoka Chakra has 24 spokes. What is the degree measure of the angle between two spokes next to each other? What is the largest acute angle formed between two spokes?

Solution

The full turn \(360^\circ\) is divided equally by the 24 spokes:

\[\text{Angle between adjacent spokes}=\frac{360^\circ}{24}=15^\circ\]

Counting spokes apart: \(5\) gaps give \(5\times15^\circ=75^\circ\) (acute), but \(6\) gaps give \(6\times15^\circ=90^\circ\) (a right angle, not acute).

Answer

Angle between two adjacent spokes \(=15^\circ\). The largest acute angle between two spokes is \(75^\circ\).

Puzzle: I am an acute angle. Doubling, tripling and quadrupling my measure still give acute angles, but multiplying by 5 gives an obtuse angle. What are the possibilities for my measure?

Solution

Let my measure be \(x\). We need:

\(4x<90^\circ\Rightarrow x<22.5^\circ\) (quadruple is still acute)
\(5x>90^\circ\Rightarrow x>18^\circ\) (five times is obtuse)

So \(18^\circ < x < 22.5^\circ\). (Also \(5x<180^\circ\Rightarrow x<36^\circ\) is automatically true.)

Answer

Any measure with \(18^\circ < x < 22.5^\circ\). In whole degrees: \(x = 19^\circ,\ 20^\circ,\ 21^\circ\) or \(22^\circ\).

Educational solutions compiled from NCERT Ganita Prakash, Grade 6 · for practice & revision · @edugrown